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            <h1 style="display: none">algorithm-初赛抱佛脚</h1>
            
              <p class="note note-info">
                
                  本文最后更新于：1 年前
                
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            <div class="markdown-body" id="post-body">
              <h2 id="前言"><a href="#前言" class="headerlink" title="前言"></a>前言</h2><p><del>该死的CCF，居然想到拿CSP当NOIP前置条件的鬼路子</del></p>
<p>在考试前一个星期甚至还夹着联考的我又又又开坑了(orz</p>
<p>抱佛脚抱佛脚抱佛脚抱佛脚</p>
<p>没啥顺序，想到啥写啥</p>
<p>语言全部以<code>cpp</code>为例</p>
<p>那种我觉得完全没必要再拿出来的东西我也没写的(小声</p>
<a id="more"></a>
<h2 id="目录"><a href="#目录" class="headerlink" title="目录"></a>目录</h2><ul>
<li>主定理</li>
<li>编码<ul>
<li>原码 反码 补码</li>
<li>ASCII</li>
</ul>
</li>
<li>逻辑运算</li>
<li>协议</li>
<li>软硬件</li>
<li><code>cpp</code>运算优先级</li>
<li>树</li>
<li>图</li>
<li>排列组合</li>
</ul>
<hr>
<h2 id="主定理"><a href="#主定理" class="headerlink" title="主定理"></a>主定理</h2><p>参考文章：<a target="_blank" rel="noopener" href="https://www.cnblogs.com/oier/p/9454539.html">❶</a></p>
<ul>
<li>主定理主要用于计算时间复杂度的递推式满足如下形式的时间复杂度计算</li>
</ul>
<script type="math/tex; mode=display">
T(n)=aT(\frac{n}{b})+f(n)</script><p>其实一般情况下，如果能够直接通过数列知识求通项的也没必要用主定理。</p>
<blockquote>
<p>前置知识：</p>
<p>$\Theta$ 严格等于，上界&amp;下界</p>
<p>$o$ 小于，上界</p>
<p>$O$ 小于等于，上界</p>
<p>$\omega$ 大于，下界</p>
<p>$\Omega$ 大于等于，下界</p>
</blockquote>
<p>定理内容：$\log n$的意思在$OI$中一般表示$\log_2 n$</p>
<p>对于$T(n)=aT(\frac{n}{b})+f(n)$</p>
<ul>
<li>当 $\exists \,\varepsilon &gt;0$ 使 $f(n) = O(n^{(\log_b{a})-\varepsilon})$ 则 $T(n) = \Theta(n^{\log_b{a}})$</li>
<li>当 $\exists \,k\ge 0$ 使 $f(n)=O(n^{\log_b{a}}\log^k n)$ 则  $T(n)=\Theta(n^{\log_b{a}}\log^{k+1} n)$</li>
<li>当 $\exists \,\varepsilon &gt; 0$ 使 $f(n) = \Omega(n^{(\log_ba)+\varepsilon})$ 同时 $\exists c&lt;1$ 以及充分大的 $n$ 满足 $af(\frac{n}{b})\leq cf(n)$ 那么 $T(n) = \Theta(f(n))$</li>
</ul>
<hr>
<h2 id="编码"><a href="#编码" class="headerlink" title="编码"></a>编码</h2><h3 id="原码-反码-补码"><a href="#原码-反码-补码" class="headerlink" title="原码 反码 补码"></a>原码 反码 补码</h3><p>参考文章：<a target="_blank" rel="noopener" href="https://www.luogu.com.cn/blog/cdcq/ExplainationOnComplement">❶</a></p>
<p><strong>只给出计算方法</strong></p>
<ul>
<li>原码：最高位表示符号的二进制编码<ul>
<li>$(0???)\text{原} = (???)_2$ </li>
<li>$(1???)\text{原}= -(???)_2$ </li>
</ul>
</li>
<li>反码：正数反码等于原码，负数反码等于除符号位各位取反  <ul>
<li>$(0???)\text{反} = (0???)\text{原} $ </li>
<li>$(1???)\text{反} = (1(!?)(!?)(!?))\text{原}$ </li>
</ul>
</li>
<li>补码：把符号位的权值设为 $-2^{w-1}$ ，再正常展开，$w$ 等于二进制长度<ul>
<li>例如：$(1111)_\text补=-1\times2^3+1\times2^2+1\times2^1+1\times2^0=-1$</li>
</ul>
</li>
</ul>
<h3 id="ASCII"><a href="#ASCII" class="headerlink" title="ASCII"></a>ASCII</h3><ul>
<li>一共有128 ($2^7$) 个(0~127)</li>
</ul>
<div class="table-container">
<table>
<thead>
<tr>
<th>字符</th>
<th>ASCII码(10进制)</th>
<th>ASCII码(16进制)</th>
</tr>
</thead>
<tbody>
<tr>
<td>‘ ‘(空格)</td>
<td>32D</td>
<td>20H</td>
</tr>
<tr>
<td>‘0’</td>
<td>48D</td>
<td>30H</td>
</tr>
<tr>
<td>‘1’</td>
<td>49D</td>
<td>31H</td>
</tr>
<tr>
<td>‘A’</td>
<td>65D</td>
<td>41H</td>
</tr>
<tr>
<td>‘a’</td>
<td>97D</td>
<td>51H</td>
</tr>
</tbody>
</table>
</div>
<hr>
<h2 id="逻辑"><a href="#逻辑" class="headerlink" title="逻辑"></a>逻辑</h2><div class="table-container">
<table>
<thead>
<tr>
<th>数学符号</th>
<th>中文表述</th>
<th>英文表述</th>
<th>集合表述</th>
<th>cpp表述</th>
</tr>
</thead>
<tbody>
<tr>
<td>$\rm \neg$</td>
<td>非</td>
<td>not</td>
<td>$\rm\complement_U$(关于集合 $\rm U$ 的补集)</td>
<td><code>!</code></td>
</tr>
<tr>
<td>$\rm\vee$</td>
<td>或</td>
<td>or</td>
<td>$\rm\cup$(并集)</td>
<td>`</td>
<td></td>
<td>`</td>
</tr>
<tr>
<td>$\rm \wedge$</td>
<td>与</td>
<td>and</td>
<td>$\rm\cap$(交集)</td>
<td><code>&amp;&amp;</code></td>
</tr>
<tr>
<td>$\rm \oplus$</td>
<td>异或</td>
<td>xor</td>
<td></td>
<td><code>^</code></td>
</tr>
</tbody>
</table>
</div>
<ul>
<li>(我也不知道xor是不是数学里的)</li>
</ul>
<p><strong>关于优先级的几点说明</strong>：</p>
<p>一般情况下都是带括号的，如果没有的话是 $\rm\neg$ -&gt; $\rm\wedge$ -&gt; $\rm\vee$</p>
<hr>
<h2 id="网络协议"><a href="#网络协议" class="headerlink" title="网络协议"></a>网络协议</h2><p>参考文章：<strong><a target="_blank" rel="noopener" href="https://www.cnblogs.com/imyalost/p/6086808.html">❶</a></strong>  <strong><a target="_blank" rel="noopener" href="https://blog.csdn.net/csdn_kou/article/details/82910753">❷</a></strong>  <strong><a href="www.baidu.com">某不公开的文件</a></strong></p>
<ul>
<li>OSI 七层模型：</li>
</ul>
<blockquote>
<p>|     应用层     | (资源子网络)</p>
<p>|     表示层     | (资源子网络)</p>
<p>|     会话层     | (资源子网络)</p>
<p>|     传输层     | (承上启下)</p>
<p>|     网络层     | (通信子网络)</p>
<p>| 数据链路层 | (通信子网络)</p>
<p>|     物理层     | (通信子网络)</p>
</blockquote>
<ul>
<li>TCP / IP 四层模型：</li>
</ul>
<blockquote>
<p>|     应用层     | <code>DNS</code> <code>ping</code> …</p>
<p>|     传输层     | <code>TCP</code> <code>UDP</code> …</p>
<p>|     网络层     | <code>IP</code> <code>ICMP</code> …</p>
<p>| 数据链路层 | </p>
</blockquote>
<ul>
<li><p>IP 地址问题</p>
<ul>
<li><code>xxx</code>.<code>xxx</code>.<code>xxx</code>.<code>xxx</code></li>
<li>每一段取值范围在<code>0~255</code></li>
</ul>
</li>
<li><p>域名</p>
</li>
</ul>
<div class="table-container">
<table>
<thead>
<tr>
<th>edu</th>
<th>com</th>
<th>net</th>
<th>info</th>
<th>org</th>
<th>ini</th>
<th>cn</th>
</tr>
</thead>
<tbody>
<tr>
<td>教育/科研机构</td>
<td>商业机构</td>
<td>网络服务机构</td>
<td>信息服务机构</td>
<td>专业团体</td>
<td>国际机构</td>
<td>中国最高层域名</td>
</tr>
</tbody>
</table>
</div>
<ul>
<li>域名管理服务 DNS</li>
</ul>
<hr>
<h2 id="软硬件"><a href="#软硬件" class="headerlink" title="软硬件"></a>软硬件</h2><p>参考文章：<strong><a target="_blank" rel="noopener" href="https://blog.csdn.net/qq_36045520/article/details/52734033">❶</a></strong></p>
<h3 id="操作系统"><a href="#操作系统" class="headerlink" title="操作系统"></a>操作系统</h3><ul>
<li>Windows</li>
<li>Unix / Linux (Ubuntu / Debian / Fedora / Elementary / Cent / RedHat / Arch / …)</li>
<li>Net ware</li>
</ul>
<h3 id="计算机内存"><a href="#计算机内存" class="headerlink" title="计算机内存"></a>计算机内存</h3><ul>
<li>ROM（只读存储器）<ul>
<li>ROM中的信息（数据或程序）一般在制造完成后就已读入并永久保存</li>
<li>ROM中的信息只能读出，不能写入</li>
<li>ROM即使断电后也不会丢失数据</li>
</ul>
</li>
<li>RAM（随机存储器）</li>
<li>RAM中的信息可以读出，也可以写入</li>
<li>RAM断电后会丢失数据</li>
</ul>
<hr>
<h2 id="cpp运算优先级"><a href="#cpp运算优先级" class="headerlink" title="cpp运算优先级"></a><code>cpp</code>运算优先级</h2><ul>
<li>【1】<code>(</code> <code>)</code> <code>[</code> <code>]</code> <code>A-&gt;x</code> <code>A.x</code> <code>++i</code> <code>--i</code></li>
<li>【2】<code>-x</code> <code>(int)/(char)/(bool)/...</code> <code>i++</code> <code>i--</code> <code>*p</code> <code>&amp;p</code> <code>!</code> <code>~</code> </li>
<li>【3】<code>/</code> <code>*</code> <code>%</code></li>
<li>【4】<code>+</code> <code>-</code></li>
<li>【5】<code>&lt;&lt;</code> <code>&gt;&gt;</code></li>
<li>【6】<code>&gt;=</code> <code>&gt;</code> <code>&lt;</code> <code>&lt;=</code></li>
<li>【7】<code>==</code> <code>!=</code> </li>
<li>【8】<code>&amp;</code></li>
<li>【9】<code>^</code></li>
<li>【10】<code>|</code></li>
<li>【11】<code>&amp;&amp;</code></li>
<li>【12】<code>||</code></li>
<li>【13】<code>+=</code> <code>-=</code> <code>/=</code> <code>*=</code> <code>%=</code> <code>&lt;&lt;=</code> <code>&gt;&gt;=</code> <code>&amp;=</code> <code>|=</code> <code>^=</code></li>
<li>【14】<code>,</code></li>
</ul>
<hr>
<h2 id="树"><a href="#树" class="headerlink" title="树"></a>树</h2><ul>
<li>$\rm n$ 个节点的二叉树能形成 $\rm \frac{C^{n}_{2n}}{2}$ 种二叉树</li>
<li>二叉树的先序\中序\后序遍历</li>
</ul>
<hr>
<h2 id="图"><a href="#图" class="headerlink" title="图"></a>图</h2><ul>
<li>欧拉图：存在<strong>欧拉回路</strong>的图<ul>
<li>欧拉回路：能回到起点的<strong>欧拉路</strong><ul>
<li>欧拉路：经过图中每一条<strong>边</strong>且只经过每条<strong>边</strong>一次的路径</li>
</ul>
</li>
</ul>
</li>
<li>哈密顿图：存在<strong>哈密顿回路</strong>的图<ul>
<li>哈密顿回路：能回到起点的<strong>哈密顿路</strong><ul>
<li>哈密顿路：经过图中每一个<strong>点</strong>且只经过每个<strong>点</strong>一次的路径</li>
</ul>
</li>
</ul>
</li>
</ul>
<hr>
<h2 id="排列组合"><a href="#排列组合" class="headerlink" title="排列组合"></a>排列组合</h2><p>感觉和高考题难度差不多吧(</p>
<p>就讲个卡特兰数(我记得以前写过blog但是好像不见了呜呜呜</p>
<ul>
<li>为了避免和组合数 $\rm C$ 混淆，这里卡特兰数用 $\rm Ca$ 表示</li>
</ul>
<script type="math/tex; mode=display">
\rm\text{递推公式：}\;\;
Ca_{n+1} = \sum_{i=0}^n Ca_{i}Ca_{n-i}</script><script type="math/tex; mode=display">
\rm\text{通项公式：}\;\;
Ca_n = \frac{1}{n+1}C^{n}_{2n}</script><p>卡特兰数的意义实际上是个匹配问题，或者说是个递推关系的应用</p>
<ul>
<li>$\rm n$ 个元素的出入栈问题</li>
<li>$\rm n$ 个括号的匹配问题</li>
<li>$\rm n\times n$ 矩阵的对角线最短格线路径问题 </li>
<li>…</li>
</ul>
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